Question

The integer n which $\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^2}} \right)}}{{{x^n}}}$ is finite non zero number is
A.1
B.2
C.3
D.4

Answer 1

Hint: First substitute the Taylor expansion series in place of $\cos x,{e^x}.$Take the first two of expansions as $x \to 0$remaining all will become zero. After substituting, find the limit in terms of n. So, try to make a condition on n to make the limit non-zero.

Complete step-by-step answer:
Given limit in the question can be written in the form of:
$\mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$
Let us assume this limit to be l. So, we get
$l = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\cos x - 1} \right)\left( {\cos x - {e^x}} \right)}}{{{x^n}}}$
Taylor expansion series: In mathematics the Taylor series of a function is an infinite sum of terms that are expressed in terms of functions derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point.
By using Taylor series we can say the cosx can be written as:
$\cos x = 1 - \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^4}}}{{4!}} + {\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}$
By using Taylor series we can say the value of ${e}^{x}$, can be said:
${e^x} = 1 + \dfrac{x}{{1!}} + \dfrac{{{x^2}}}{{2!}}$
Now substitute this into the equation of l, we get:
$l = \dfrac{{\mathop {\lim }\limits_{x \to 0} \left( {1 - \dfrac{{{x^2}}}{{2!}} - 1} \right)\left( {1 - \dfrac{{{x^2}}}{{2!}} - \left( {1 + \dfrac{x}{{1!}}} \right)} \right)}}{{{x^n}}}$
We took only 2 terms because at higher power x $x \to 0$it becomes 0.
By simplifying we can write the value of l as follows:
$l = \dfrac{{\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{x^2}}}{{2!}}} \right)\left( {1 - \dfrac{{{x^2}}}{{2!}} - 1 - \dfrac{x}{{1!}}} \right)}}{{{x^n}}}$
By simplifying we can write the value of l as follows:
$l = \dfrac{{\mathop {\lim }\limits_{x \to 0} {\rm{ }}\left( {\dfrac{{{x^2}}}{{2!}}} \right)\left( { - \dfrac{x}{{1!}} - \dfrac{{{x^2}}}{{2!}}} \right)}}{{{x^n}}}$
We can neglect ${x^2}$ terms in the second bracket as x is present.
$l = \dfrac{{\mathop {\lim }\limits_{x \to 0} \left( { - \dfrac{{{x^2}}}{{2!}}} \right)\left( { - \dfrac{x}{{1!}}} \right)}}{{{x^n}}}$
We can simplify it to the extent of:
$l = \mathop {\lim }\limits_{x \to 0} \dfrac{{{x^3}}}{{2{x^n}}}$
So, if n=3 then ${{\rm{x}}^{\rm{3}}}$term cancels to give limit $\dfrac{1}{2}$
But if ${\rm{n}} < {\rm{3}}$ then $\lim = 0$, if ${\rm{n}} > {\rm{3}}$ limit does not exist.
So, $n = 3$is the answer.
Therefore, option (c) is correct.

Note: Don’t confuse where to neglect and where to not neglect. You must take least degree n in both the terms since when you multiply you went least degree term as the other terms tend to zero. The condition between n, l is very important and conclude it with at most care.